Consider the following half cell reaction
Cr₂O₇²⁻ (aq) + 6e⁻ + 14H⁺ (aq) → 2Cr³⁺ (aq) + 7H₂O (ℓ)
The reaction was conducted with the ratio of [Cr³⁺]²/[Cr₂O₇²⁻] = 10⁻⁶. The pH value at which the EMF of the half cell will become zero is __________.
(nearest integer value)
[Given: standard half cell reduction potential E°_(Cr₂O₇²⁻, H⁺/Cr³⁺) = 1.33 V, 2.303RT/F = 0.059 V.]

Solution

Chemistry ma’am · 4 steps

Set E = 0 in the Nernst equation and let [H⁺]¹⁴ carry the whole pH dependence.

1. Write Nernst. For this reduction n = 6, and H⁺ appears in the reaction quotient with power 14. Water is a liquid, so it stays out of Q.

   E = E° - 0.059/6 log[Cr³⁺]²/[Cr₂O₇²⁻][H⁺]¹⁴

2. Set E to zero. Then log Q = (6 × 1.33)/0.059 = 7.98/0.059 = 135.25.
3. Bring in pH. log Q = log 10⁻⁶ - 14 log[H⁺] = -6 + 14 pH. So 14 pH = 135.25 + 6 = 141.25.
4. So. pH = 141.25/14 = 10.09, which rounds to 10.

The common slip: 9. Moving the -6 to the wrong side gives 14 pH = 135.25 - 6 = 129.25, so pH ≈ 9, not 10.

The idea to keep. When H⁺ sits in a half-reaction, -log[H⁺]ᵐ becomes m pH. Write log Q as known part plus m pH, then solve.

Answer from the NTA final answer key. Solution by PrepWiser.